Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

const int INF=1001001001001001001ll;

struct segtree{
    static const int SEG=1<<19;
    vint dat,put;
    segtree():dat(SEG*2,INF),put(SEG*2,INF){}
    inline void push(int k){
        chmin(dat[k],put[k]);
        if(k<SEG-1){
            chmin(put[k*2+1],put[k]);
            chmin(put[k*2+2],put[k]);
        }
    }
    void update(int a,int b,int x,int k=0,int l=0,int r=SEG){
        push(k);
        if(r<=a||b<=l)return;
        if(a<=l&&r<=b){
            put[k]=x;
            push(k);
            return;
        }
        update(a,b,x,k*2+1,l,(l+r)/2);
        update(a,b,x,k*2+2,(l+r)/2,r);
        dat[k]=min(dat[k*2+1],dat[k*2+2]);
    }
    int query(int a,int b,int k=0,int l=0,int r=SEG){
        push(k);
        if(r<=a||b<=l)return INF;
        if(a<=l&&r<=b)return dat[k];
        return min(query(a,b,k*2+1,l,(l+r)/2),query(a,b,k*2+2,(l+r)/2,r));
    }
};

int N,K;
int A[111111],B[111111];
int S[111111];

int dp[111111];
signed main(){
    cin>>N>>K;
    rep(i,N){
        cin>>A[i]>>B[i];
        A[i]*=2;
    }
    rep(i,N){
        if(A[i]>K&&B[i]==1){
            cout<<-1<<endl;
            return 0;
        }
    }

    rep(i,N)S[i+1]=S[i]+A[i];
    vint lis;
    rep(i,N+1){
        lis.pb(S[i]%K);
    }
    sort(all(lis));lis.erase(unique(all(lis)),lis.end());
    segtree seg;
    for(int i=N-1;i>=0;i--){
        int w=lower_bound(all(lis),S[i]%K)-lis.begin();
        int z=seg.query(w,w+1);
        if(z==INF){
            dp[i]=S[N]-S[i];
        }
        else{
            dp[i]=((S[z]-S[i])/K+1)*K+dp[z];
        }
        if(B[i]==1){
            int l=lower_bound(all(lis),S[i]%K)-lis.begin();
            int r=lower_bound(all(lis),S[i+1]%K)-lis.begin();
            if(l<r){
                seg.update(l+1,r,i);
            }
            else{
                seg.update(l+1,lis.size(),i);
                seg.update(0,r,i);
            }
        }
    }

    rep(i,lis.size()){
        if(seg.query(i,i+1)==INF){
            cout<<S[N]<<endl;
            return 0;
        }
    }

    seg=segtree();

    int ans=INF;
    rep(i,N){
        int w=lower_bound(all(lis),S[i]%K)-lis.begin();
        if(seg.query(w,w+1)==INF){
            chmin(ans,S[i]+dp[i]);
        }

        if(B[i]==1){
            int l=lower_bound(all(lis),S[i]%K)-lis.begin();
            int r=lower_bound(all(lis),S[i+1]%K)-lis.begin();
            if(l<r){
                seg.update(l+1,r,i);
            }
            else{
                seg.update(l+1,lis.size(),i);
                seg.update(0,r,i);
            }
        }
    }

    cout<<ans<<endl;
    return 0;
}